Let's go through the first question (Q1) step by step, as requested. I will solve each differential equation listed under Q1. --- ### **Q1. Find a general solution. Show the steps.** #### **(a) \( y^3 y' + x^3 = 0 \)** **Step 1: Rewrite the equation** \[ y^3 \frac{dy}{dx} = -x^3 \] **Step 2: Separate variables** \[ y^3 dy = -x^3 dx \] **Step 3: Integrate both sides** \[ \int y^3 dy = \int -x^3 dx \] \[ \frac{y^4}{4} = -\frac{x^4}{4} + C \] **Step 4: Solve for the general solution** \[ y^4 + x^4 = 4C \] Or, let \(C' = 4C\): \[ \boxed{y^4 + x^4 = C} \] **Check by substitution:** Differentiate both sides implicitly: \[ 4y^3 y' + 4x^3 = 0 \implies y^3 y' + x^3 = 0 \] --- #### **(b) \( y' \sin 2\pi x = \pi y \cos 2\pi x \)** **Step 1: Rearrange and separate variables** \[ y' = \frac{\pi y \cos 2\pi x}{\sin 2\pi x} \] \[ \frac{dy}{dx} = \pi y \cot 2\pi x \] \[ \frac{dy}{y} = \pi \cot 2\pi x \, dx \] **Step 2: Integrate both sides** \[ \int \frac{dy}{y} = \pi \int \cot 2\pi x \, dx \] \[ \ln |y| = \pi \cdot \frac{1}{2\pi} \ln |\sin 2\pi x| + C \] \[ \ln |y| = \frac{1}{2} \ln |\sin 2\pi x| + C \] **Step 3: Exponentiate** \[ |y| = D |\sin 2\pi x|^{1/2} \] where \(D = e^C\). \[ \boxed{y = C \sqrt{\sin 2\pi x}} \] --- #### **(c) \( yy' + 36x = 0 \)** **Step 1: Write as \(y \frac{dy}{dx} = -36x\)** \[ y \frac{dy}{dx} = -36x \] **Step 2: Separate variables** \[ y dy = -36x dx \] **Step 3: Integrate** \[ \int y dy = \int -36x dx \] \[ \frac{y^2}{2} = -18x^2 + C \] **Step 4: General solution** \[ y^2 + 36x^2 = 2C \] Or, \[ \boxed{y^2 + 36x^2 = C} \] --- #### **(d) \( y' = e^{2x} - y^2 \)** This is a **Riccati equation** (nonlinear), but can be solved by separation if rewritten. \[ \frac{dy}{dx} = e^{2x} - y^2 \] This is not separable directly. Let's try substitution. Let \(y = v(x)\). \[ \frac{dy}{dx} + y^2 = e^{2x} \] This is a Bernoulli equation. Let's try \(n=2\): Let \(y = \frac{1}{u}\), so \(\frac{dy}{dx} = -\frac{1}{u^2} \frac{du}{dx}\): \[ -\frac{1}{u^2} \frac{du}{dx} + \frac{1}{u^2} = e^{2x} \] \[ -\frac{du}{dx} + 1 = e^{2x} u^2 \] But this is more complex and requires advanced techniques (special integrating factor or substitution). This is typically left in implicit form unless further guidance is given. --- ## **Summary of General Solutions (Q1):** **(a)**  \(\boxed{y^4 + x^4 = C}\) **(b)**  \(\boxed{y = C \sqrt{\sin 2\pi x}}\) **(c)**  \(\boxed{y^2 + 36x^2 = C}\) **(d)**  Non-elementary solution; nonlinear ODE (leave as is or solve with advanced techniques). --- Would you like solutions for Q2, Q3, and Q4 as well?

Certainly! Let's proceed with the solutions for Q2, Q3, and Q4, step by step. --- ### **Q2. Solve the IVP:** \[ \frac{dy}{dt} = 2t y, \quad y(1) = 3 \] **Step 1: Find the general solution** This is a first-order linear ODE: \[ \frac{dy}{dt} = 2 t y \] **Step 2: Separate variables** \[ \frac{dy}{y} = 2 t dt \] **Step 3: Integrate both sides** \[ \int \frac{dy}{y} = \int 2 t dt \] \[ \ln |y| = t^2 + C \] **Step 4: Write the general solution** \[ |y| = e^{t^2 + C} = Ae^{t^2} \] where \(A = e^{C}\). **Step 5: Apply initial condition \(y(1) = 3\)** \[ 3 = A e^{1^2} = A e \] \[ A = \frac{3}{e} \] **Final particular solution:** \[ \boxed{ \boxed{y(t) = \frac{3}{e} e^{t^2} = 3 e^{t^2 - 1} } \] --- ### **Q3. Test for exactness and solve if exact** --- #### **(a) \( 2x y dx + x^2 dy = 0 \)** Rewrite: \[ M(x,y) = 2x y, \quad N(x,y) = x^2 \] **Step 1: Check if exact** \[ \frac{\partial M}{\partial y} = 2x \] \[ \frac{\partial N}{\partial x} = 2x \] Since \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} = 2x\), **the equation is exact**. **Step 2: Find potential function \(\Psi(x,y)\)** \[ \frac{\partial \Psi}{\partial x} = M = 2x y \] Integrate w.r.t. \(x\): \[ \Psi(x,y) = x^2 y + h(y) \] Differentiate w.r.t. \(y\): \[ \frac{\partial \Psi}{\partial y} = x^2 + h'(y) \] Set equal to \(N = x^2\): \[ x^2 + h'(y) = x^2 \Rightarrow h'(y) = 0 \] \[ h(y) = \text{constant} \] **Solution:** \[ \boxed{ x^2 y = C } \] --- #### **(b) \( 2x \tan y dx + \sec^2 y dy = 0 \)** Rewrite: \[ M = 2x \tan y, \quad N = \sec^2 y \] **Step 1: Check if exact** \[ \frac{\partial M}{\partial y} = 2x \sec^2 y \] \[ \frac{\partial N}{\partial x} = 0 \] Since these are not equal, **not exact**. **Step 2: Find an integrating factor** Observe that \(\frac{\partial N}{\partial x} = 0\), so the equation can be made exact by multiplying through by an integrating factor depending on \(y\). Divide through by \(\sec^2 y\): \[ 2x \tan y \cdot \frac{1}{\sec^2 y} dx + dy = 0 \] But perhaps a better approach is to note that: \[ \frac{\partial}{\partial y} (2x \ln |\sec y|) = 2x \tan y \] Alternatively, recognize that the equation simplifies if we write: \[ \frac{dy}{dx} = - \frac{2x \tan y}{\sec^2 y} = -2x \tan y \cdot \cos^2 y \] Since \(\tan y \cdot \cos^2 y = \sin y \cos y\): \[ \frac{dy}{dx} = -2x \sin y \cos y \] Now, this is separable: \[ \frac{dy}{\sin y \cos y} = -2x dx \] Recall: \[ \frac{1}{\sin y \cos y} = \frac{2}{\sin 2 y} \] So: \[ \frac{dy}{\sin y \cos y} = \frac{2 dy}{\sin 2 y} \] Thus: \[ \frac{2 dy}{\sin 2 y} = -2x dx \] \[ \frac{dy}{\sin 2 y} = -x dx \] **Step 3: Integrate both sides** Let \(u = 2 y\), so \(du = 2 dy\): \[ \frac{dy}{\sin 2 y} = \frac{1}{2} \frac{du}{\sin u} \] Therefore: \[ \frac{1}{2} \int \frac{du}{\sin u} = - \int x dx \] Recall: \[ \int \frac{du}{\sin u} = \int \csc u \, du = - \ln |\csc u + \cot u| + C \] So: \[ \frac{1}{2} \left( - \ln |\csc u + \cot u| \right) = - \frac{x^2}{2} + C \] \[ - \frac{1}{2} \ln |\csc 2 y + \cot 2 y| = - \frac{x^2}{2} + C \] Multiply both sides by -2: \[ \ln |\csc 2 y + \cot 2 y| = x^2 + C' \] Exponentiate: \[ |\csc 2 y + \cot 2 y| = K e^{x^2} \] --- ### **Particular solution (if initial condition given):** If initial condition is provided, solve for \(K\). Otherwise, the implicit solution is: \[ \boxed{ \csc 2 y + \cot 2 y = \pm e^{x^2 + C'} } \] --- ### **(c) \( x y dx + dy = 0 \), with \( y(0) = 2 \)** Rewrite: \[ x y dx + dy = 0 \] \[ dy = - x y dx \] **Step 1: Separate variables** \[ \frac{dy}{y} = - x dx \] **Step 2: Integrate** \[ \int \frac{dy}{y} = - \int x dx \] \[ \ln |y| = - \frac{x^2}{2} + C \] **Step 3: Apply initial condition \( y(0) = 2 \)** \[ \ln 2 = - 0 + C \Rightarrow C = \ln 2 \] **Solution:** \[ \ln | y | = - \frac{x^2}{2} + \ln 2 \] \[ | y | = 2 e^{- x^2/2} \] Since \( y(0) = 2 > 0 \): \[ \boxed{ \boxed{ y(x) = 2 e^{- x^2/2} } } \] --- ### **Q4. Find the general solution:** --- #### **(a) \( y - y' = 52 \)** Rewrite: \[ y' - y = -52 \] This is a linear first-order ODE. **Step 1: Integrating factor** \[ \mu(x) = e^{-\int 1 dx} = e^{-x} \] **Step 2: Multiply through** \[ e^{-x} y' - e^{-x} y = -52 e^{-x} \] Left side is derivative: \[ \frac{d}{dx} [ y e^{-x} ] = -52 e^{-x} \] **Step 3: Integrate** \[ y e^{-x} = \int -52 e^{-x} dx + C \] \[ = -52 \int e^{-x} dx + C = -52 (- e^{-x}) + C = 52 e^{-x} + C \] **Step 4: Solve for \( y \)** \[ y e^{-x} = 52 e^{-x} + C \] \[ y = 52 + C e^{x} \] --- #### **(b) \( y = 2 y - 4 x \)** Assuming the equation is: \[ y = 2 y - 4 x \] Rearranged: \[ 0 = y - 4 x \] \[ \boxed{ y = 4 x } \] which is a particular solution. --- #### **(c) \( y + ky = e^{-k x} \)** Assuming the equation is: \[ (1 + k) y = e^{-k x} \] \[ \boxed{ y = \frac{e^{-k x}}{1 + k} } \] or, if the original was a differential equation involving \( y \), please clarify. --- **Summary of Q4 solutions:** - (a) \( y = 52 + C e^{x} \) - (b) \( y = 4 x \) - (c) \( y = \frac{e^{-k x}}{1 + k} \) --- ### **Final notes:** If you have specific initial conditions or need further clarification, please let me know!